Question

In the given figure, LMN is tangent to the circle with centre O. If $\angle$ PMN = ${60}^{\circ }$, find $\angle$MOP.

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is D

${120}^{\circ }$

$\angle$OMN = ${90}^{\circ }$ (Radius and tangent are perpendicular at the point of contact)
$\angle OMP=\angle$OMN - $\angle$PMN
$\angle OMP={90}^{\circ }-{60}^{\circ }={30}^{\circ }$
OP = OM = Radius
Hence, $\angle$OMP = $\angle$OPM = ${30}^{\circ }$

$\angle$OMP + $\angle$OPM + $\angle$MOP = ${180}^{\circ }$ (Angle sum property)
$\angle$MOP = ${180}^{\circ }$ - $\angle$OMP - $\angle$OPM
$\angle$MOP = ${180}^{\circ }$ - ${30}^{\circ }$ - ${30}^{\circ }$ = ${120}^{\circ }$.